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Let f: R--> R be a function defined by f(x) = min{x + 1,|x|+1}; then prove that f(x) is diffrentiable everywhere
Dear Meghana This can be seen as problem if arise will be due to modulus function i.e. |x| + 1 Now it will be non differentiable at x= 0 but at x = 0 x+ 1 will be taken as main function. hence f(x) is diffrentiable everywhere
the correct answer would be f(x)= x+1 (because |x| greater than equal to x for all x)so f’(x)=1 and hence it is differentiable everywhere
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